A048739 -id:A048739 - cp's OEIS frontend

A082574 a(1)=1, a(n) = ceiling(r(3)a(n-1)) where r(3) = (1/2)(3 + sqrt(13)) is the positive root of X^2 = 3*X + 1.

1, 4, 14, 47, 156, 516, 1705, 5632, 18602, 61439, 202920, 670200, 2213521, 7310764, 24145814, 79748207, 263390436, 869919516, 2873148985, 9489366472, 31341248402, 103513111679, 341880583440, 1129154862000, 3729345169441, 12317190370324

Offset: 1

Benoit Cloitre, May 06 2003

nonn

More generally the sequence a(1)=1, a(n) = ceiling(r(z)*a(n-1)) where r(z) = (1/2)*(z + sqrt(z^2 + 4)) is the positive root of X^2 = z*X + 1 satisfies the linear recurrence: for n > 3, a(n) = (z+1)*a(n-1) - (z-1)*a(n-2) - a(n-3) and the closed-form formula: a(n) = floor(t(z)*r(z)^n) where t(z) = (1/(2*z))*(1+(z+2)/sqrt(z^2+4)) is the positive root of z*(z^2 + 4)*X^2 = (z^2 + 4)*X + 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Ignas Gasparavičius, Andrius Grigutis, and Juozas Petkelis, Picturesque convolution-like recurrences and partial sums' generation, arXiv:2507.23619 [math.NT], 2025. See p. 24.
Index entries for linear recurrences with constant coefficients, signature (4,-2,-1).

Cf. A000071, A048739, A049652.

I:=[1,4,14]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 12 2017

a:=n->sum(fibonacci(i,3), i=0..n): seq(a(n), n=1..30); # Zerinvary Lajos, Mar 20 2008

LinearRecurrence[{4, -2, -1}, {1, 4, 14}, 30] (* Vincenzo Librandi, Sep 12 2017 *)
Table[Sum[Fibonacci[k, 3], {k,0,n}], {n,1,30}] (* G. C. Greubel, May 31 2019 *)

Vec(1/((1-x)*(1-3*x-x^2)) + O(x^30)) \\ Michel Marcus, Sep 12 2017

(1/((1-x)*(1-3*x-x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 31 2019

a(1)=1, a(2)=4, a(3)=14, a(n) = 4*a(n-1) - 2*a(n-2) - a(n-3).
a(n) = floor(t(3)*r(3)^n) where t(3) = (1/6)*(1 + 5/sqrt(13)) is the positive root of 39*X^2 = 13*X + 1.
G.f.: 1/((1-x)*(1-3*x-x^2)). Partial sums of A006190. - Paul Barry, Jul 10 2004

A096979 Sum of the areas of the first n+1 Pell triangles.

Original entry on oeis.org

0, 1, 6, 36, 210, 1225, 7140, 41616, 242556, 1413721, 8239770, 48024900, 279909630, 1631432881, 9508687656, 55420693056, 323015470680, 1882672131025, 10973017315470, 63955431761796, 372759573255306, 2172602007770041

Offset: 0

Paul Barry, Jul 17 2004

Convolution of A059841(n) and A001109(n+1).

Partial sums of A084158.

S. Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Index entries for linear recurrences with constant coefficients, signature (6,0,-6,1).

Cf. A096977, A064831, A096978.

Accumulate[LinearRecurrence[{5,5,-1},{0,1,5},30]] (* Harvey P. Dale, Sep 07 2011 *)
LinearRecurrence[{6, 0, -6, 1},{0, 1, 6, 36},22] (* Ray Chandler, Aug 03 2015 *)

G.f.: x/((1-x)*(1+x)*(1-6*x+x^2)).
a(n) = 6*a(n-1)-6*a(n-3)+a(n-4).
a(n) = (3-2*sqrt(2))^n*(3/32-sqrt(2)/16)+(3+2*sqrt(2))^n*(sqrt(2)/16+3/32)-(-1)^n/16-1/8.
a(n) = Sum_{k=0..n} (sqrt(2)*(sqrt(2)+1)^(2*k)/8-sqrt(2)*(sqrt(2)-1)^(2*k)/8)*(1+(-1)^(n-k))/2.
a(n) = Sum_{k=0..n} A000129(k)*A000129(k+1)/2. [corrected by Jason Yuen, Jan 14 2025]
a(n) = (A001333(n+1)^2 - 1)/8 = ((A000129(n) + A000129(n+1))^2 - 1)/8. - Richard R. Forberg, Aug 25 2013
a(n) = A002620(A000129(n+1)) = A000217(A048739(n-1)), n > 0. - Ivan N. Ianakiev, Jun 21 2014

A193845 Mirror of the triangle A193844.

Original entry on oeis.org

1, 3, 1, 7, 5, 1, 15, 17, 7, 1, 31, 49, 31, 9, 1, 63, 129, 111, 49, 11, 1, 127, 321, 351, 209, 71, 13, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Clark Kimberling, Aug 07 2011

This triangle is obtained by reversing the rows of the triangle A193844.
From Philippe Deléham, Jan 17 2014: (Start)
Subtriangle of the triangle in A112857.
T(n,0)   = A000225(n+1).
T(n,1)   = A000337(n).
T(n+2,2) = A055580(n).
T(n+3,3) = A027608(n).
T(n+4,4) = A211386(n).
T(n+5,5) = A211388(n).
T(n,n)   = A000012(n).
T(n+1,n) = A005408(n).
T(n+2,n) = A056220(n+2).
T(n+3,n) = A199899(n+1).
Row sums are A003462(n+1).
Diagonal sums are A048739(n).
Riordan array (1/((1-2*x)*(1-x)), x/(1-2*x)). (End)
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-2)^0 + A_1*(x-2)^1 + A_2*(x-2)^2 + ... + A_n*(x-2)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
The n-th row lists the coefficients of the polynomial sum_{k=0..n} (X+2)^k, in order of increasing powers. - M. F. Hasler, Oct 15 2014

			First six rows:
1
3....1
7....5....1
15...17...7....1
31...49...31...9...1
63...129..111..49..11..1

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
Russell Jay Hendel, A Method for Uniformly Proving a Family of Identities, arXiv:2107.03549 [math.CO], 2021.

Cf. A193844.
Cf. Columns: A000225, A000337, A055580, A027608, A211386, A211388
Cf. Diagonals: A000012, A005408, A056220, A199899

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)
Table[2^k*Binomial[n + 1, k]*Hypergeometric2F1[1, -k, -k + n + 2, 1/2], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Nov 09 2021 *)

for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(2^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ Derek Orr, Oct 14 2014

T(n,k) = A193844(n,n-k).

T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014

A210197 Triangle of coefficients of polynomials u(n,x) jointly generated with A210198; see the Formula section.

Original entry on oeis.org

1, 3, 7, 1, 15, 5, 31, 17, 1, 63, 49, 7, 127, 129, 31, 1, 255, 321, 111, 9, 511, 769, 351, 49, 1, 1023, 1793, 1023, 209, 11, 2047, 4097, 2815, 769, 71, 1, 4095, 9217, 7423, 2561, 351, 13, 8191, 20481, 18943, 7937, 1471, 97, 1, 16383, 45057, 47103

Offset: 1

Clark Kimberling, Mar 18 2012

Column 1:  -1+2^n
Row sums:  A048739
Alternating row sums: triangular numbers, A000217
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
3
7....1
15...5
31...17...1
First three polynomials u(n,x): 1, 3, 7 + x.

K. Dilcher, K. B. Stolarsky, Nonlinear recurrences related to Chebyshev polynomials, The Ramanujan Journal, 2014, Online Oct. 2014, pp. 1-23. See Table 1.

Cf. A210198, A208510.

Essentially the same as the triangle in A257597.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A210197 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210198 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A048739 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A005409 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A000217 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A000027 *)

u(n,x)=u(n-1,x)+v(n-1,x)+1,
v(n,x)=(x+1)*u(n-1,x)+v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A293004 Expansion of 2*x^2 / (x^3 + x^2 - 3x + 1).

Original entry on oeis.org

0, 0, 2, 6, 16, 40, 98, 238, 576, 1392, 3362, 8118, 19600, 47320, 114242, 275806, 665856, 1607520, 3880898, 9369318, 22619536, 54608392, 131836322, 318281038, 768398400, 1855077840, 4478554082, 10812186006, 26102926096, 63018038200, 152139002498

Offset: 0

J. Devillet, Sep 28 2017

Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one maximal element for the weak ordering R.

Colin Barker, Table of n, a(n) for n = 0..1000
Andrei Asinowski, Cyril Banderier, Sara Billey, Benjamin Hackl, Svante Linusson, Pop-stack sorting and its image: permutations with overlapping runs, Eurocomb, Acta Math. Univ. Comenianae (2019), 1-8.
M. Couceiro, J. Devillet, and J.-L. Marichal, Quasitrivial semigroups: characterizations and enumerations, arXiv:1709.09162 [math.RA], 2017.
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A000129, A048739.

Essentially the same as A265278.

I:=[0,0,2]; [n le 3 select I[n] else 3*Self(n-1)-Self(n-2)-Self(n-3): n in [1..40]]; // Vincenzo Librandi, Oct 09 2017

A293004:=gfun:-rectoproc({a(n)=3*a(n-1) -a(n-2)-a(n-3),a(0)=0,a(1)=0,a(2)=2},a(n),remember):  map(A293004, [$0..10^3]);  # Muniru A Asiru, Oct 09 2017

CoefficientList[Series[2 x^2/(x^3 + x^2 - 3 x + 1), {x, 0, 30}], x] (* Michael De Vlieger, Oct 06 2017 *)
RecurrenceTable[{a[1]==a[2]==0, a[3]==2, a[n]==3a[n-1] - a[n-2] - a[n-3]}, a, {n, 40}] (* Vincenzo Librandi, Oct 09 2017 *)

concat(vector(2), Vec(2*x^2 / (x^3+x^2-3*x+1) + O(x^40))) \\ Colin Barker, Sep 28 2017

G.f.: 2*x^2 / (x^3 + x^2 - 3x + 1).
a(n) = 2*A048739(n-2), a(0) = a(1) = 0.
From Colin Barker, Sep 28 2017: (Start)
a(n) = 3*a(n-1) - a(n-2) - a(n-3) for n > 2.
a(n) = (-2 + (1-sqrt(2))^n + (1+sqrt(2))^n) / 2. (End)
a(n) = A265278(n) for n != 1. - Joerg Arndt, Oct 01 2017

A077921 Expansion of (1-x)^(-1)/(1+2*x-x^2).

Original entry on oeis.org

1, -1, 4, -8, 21, -49, 120, -288, 697, -1681, 4060, -9800, 23661, -57121, 137904, -332928, 803761, -1940449, 4684660, -11309768, 27304197, -65918161, 159140520, -384199200, 927538921, -2239277041, 5406093004, -13051463048, 31509019101, -76069501249, 183648021600

Offset: 0

N. J. A. Sloane, Nov 17 2002

Partial sums of sequence of signed Pell numbers (-1)^n*A000129(n). - Paul Barry, May 09 2003

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (-1,3,-1).

-a(-3-n) = A048739(n).

CoefficientList[Series[(1/(1-x))/(1+2x-x^2), {x,0,50}], x] (* Harvey P. Dale, Mar 20 2011 *)

Vec((1-x)^(-1)/(1+2*x-x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

G.f.: 1/((1-x)*(1+2*x-x^2)).
From Colin Barker, Apr 15 2016: (Start)
a(n) = -a(n-1)+3*a(n-2)-a(n-3) for n>2.
a(n) = (2-(-1-sqrt(2))^(1+n)-(-1+sqrt(2))^(1+n))/4.
(End)
E.g.f.: (1/4)*(2*exp(x) + (1 + sqrt(2))*exp((-1-sqrt(2))*x) - (sqrt(2) - 1)*exp((sqrt(2)-1)*x)). - Ilya Gutkovskiy, Apr 15 2016

A210198 Triangle of coefficients of polynomials v(n,x) jointly generated with A210197; see the Formula section.

Original entry on oeis.org

1, 3, 1, 7, 4, 15, 12, 1, 31, 32, 6, 63, 80, 24, 1, 127, 192, 80, 8, 255, 448, 240, 40, 1, 511, 1024, 672, 160, 10, 1023, 2304, 1792, 560, 60, 1, 2047, 5120, 4608, 1792, 280, 12, 4095, 11264, 11520, 5376, 1120, 84, 1, 8191, 24576, 28160, 15360, 4032

Offset: 1

Clark Kimberling, Mar 18 2012

Row sums:  A005409
Column 1:  -1+2^n
Alternating row sums: 1, 2,3,4,5,6,..., A000027
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
3....1
15...12...1
31...32...6
63...80...24...1
First three polynomials v(n,x): 1, 3 + x , 15 + 12x + x^2.

Cf. A210197, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A210197 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210198 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A048739 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A005409 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A000217 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A000027 *)

u(n,x)=u(n-1,x)+v(n-1,x)+1,
v(n,x)=(x+1)*u(n-1,x)+v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A210595 Triangle of coefficients of polynomials v(n,x) jointly generated with A209999; see the Formula section.

Original entry on oeis.org

1, 2, 1, 3, 3, 2, 4, 6, 7, 3, 5, 10, 16, 13, 5, 6, 15, 30, 35, 25, 8, 7, 21, 50, 75, 76, 46, 13, 8, 28, 77, 140, 181, 157, 84, 21, 9, 36, 112, 238, 371, 413, 317, 151, 34, 10, 45, 156, 378, 686, 924, 911, 625, 269, 55, 11, 55, 210, 570, 1176, 1848, 2206, 1949, 1211, 475, 89

Offset: 1

Clark Kimberling, Mar 23 2012

Row n starts with n and ends with F(n), where F=A000045 (Fibonacci numbers).
Row sums: A048739.
Alternating row sums: 1,1,2,2,3,3,4,4,5,5, ...
For a discussion and guide to related arrays, see A208510.

			First few rows are:
  1;
  2,  1;
  3,  3,  2;
  4,  6,  7,  3;
  5, 10, 16, 13,  5;
  6, 15, 30, 35, 25,  8;
  7, 21, 50, 75, 76, 46, 13;
First few polynomials v(n,x) are:
  v(1, x) = 1;
  v(2, x) = 2 +  1*x;
  v(3, x) = 3 +  3*x +  2*x^2;
  v(4, x) = 4 +  6*x +  7*x^2 +  3*x^3;
  v(5, x) = 5 + 10*x + 16*x^2 + 13*x^3 + 5*x^4;

G. C. Greubel, Rows n = 1..30 of the triangle, flattened

Cf. A208510, A210565.

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 16;
u[n_, x_]:= x*u[n-1, x] + (1+x)*v[n-1, x] + 1;
v[n_, x_]:= x*u[n-1, x] + v[n-1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A210565 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210595 *)
(* Second program *)
v[n_, x_]:= v[n, x]= If[n<2, n+1 +n*x, (1+x)*v[n-1, x] +x^2*v[n-2, x] +1];
T[n_]:= CoefficientList[Series[v[n, x], {x,0,n}], x];
Table[T[n-1], {n, 12}]//Flatten (* G. C. Greubel, May 24 2021 *)

@CachedFunction
def v(n,x): return n+1+n*x if (n<2) else (1+x)*v(n-1,x) +x^2*v(n-2,x) +1
def T(n): return taylor( v(n,x) , x,0,n).coefficients(x, sparse=False)
flatten([T(n-1) for n in (1..12)]) # G. C. Greubel, May 24 2021

u(n,x) = x*u(n-1,x) + (x+1)*v(n-1,x) + 1,
v(n,x) = x*u(n-1,x) + v(n-1,x) + 1,
where u(1,x) = 1, v(1,x) = 1.
T(n, k) = [x^k]( v(n,x) ), where v(n, x) = (1+x)*v(n-1, x) + x^2*v(n-2, x) + 1, v(1, x) = 1, and v(2, x) = 2 + x. - G. C. Greubel, May 24 2021

A048772 Partial sums of A048696.

Original entry on oeis.org

1, 10, 29, 76, 189, 462, 1121, 2712, 6553, 15826, 38213, 92260, 222741, 537750, 1298249, 3134256, 7566769, 18267802, 44102381, 106472572, 257047533, 620567646, 1498182833

Offset: 0

Barry E. Williams

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A001333, A000129, A048694-A048696, A105082.

a048772 n = a048772_list !! n
a048772_list = scanl1 (+) a048696_list
-- Reinhard Zumkeller, Dec 15 2013

Accumulate[LinearRecurrence[{2,1},{1,9},30]] (* or *) LinearRecurrence[ {3,-1,-1},{1,10,29},30] (* Harvey P. Dale, Apr 20 2012 *)

a(n)=([0,1,0; 0,0,1; -1,-1,3]^n*[1;10;29])[1,1] \\ Charles R Greathouse IV, Feb 10 2017

a(n)=2*a(n-1)+a(n-2)+8; a(0)=1, a(1)=10.
a(n)=[ {(9+5*sqrt(2))(1+sqrt(2))^n - (9-5*sqrt(2))(1-sqrt(2))^n}/2*sqrt(2) ]-4.
a(0)=1, a(1)=10, a(2)=29, a(n)=3*a(n-1)-a(n-2)-a(n-3). - Harvey P. Dale, Apr 20 2012
G.f. ( 1+7*x ) / ( (x-1)*(x^2+2*x-1) ). a(n)=A048739(n)+7*A048739(n-1). - R. J. Mathar, Nov 08 2012

A153346 Triangle read by rows: A000012 * A153345.

Original entry on oeis.org

1, 3, 0, 7, 1, 0, 14, 5, 1, 0, 26, 16, 6, 1, 0, 46, 42, 23, 7, 1, 0, 79, 98, 71, 31, 8, 1, 0, 133, 212, 192, 109, 40, 9, 1, 0, 221, 435, 475, 332, 157, 50, 10, 1, 0, 364, 859, 1102, 916, 529, 216, 61, 11, 1, 0, 596, 1648, 2436, 2350, 1602, 795, 287, 73, 12, 1, 0, 972, 3092, 5186, 5702, 4481, 2613, 1143, 371, 86, 13, 1, 0

Offset: 0

Gary W. Adamson, Dec 24 2008

Row sums = A048739: (1, 3, 8, 20, 49, 119, 288, ...).

Left border = A001924.

			First few rows of the triangle:
    1;
    3,    0;
    7,    1,    0;
   14,    5,    1,    0;
   26,   16,    6,    1,    0;
   46,   42,   23,    7,    1,   0
   79,   98,   71,   31,    8,   1,   0;
  133,  212,  192,  109,   40,   9,   1,  0;
  221,  435,  475,  332,  157,  50,  10,  1,  0;
  364,  859, 1102,  916,  529, 216,  61, 11,  1, 0;
  596, 1648, 2436, 2350, 1602, 795, 287, 73, 12, 1, 0;
  ...

Cf. A001924, A048739, A153345.

a(44) = 0 corrected and more terms from Georg Fischer, Jun 05 2023

cp's OEIS Frontend

A082574 a(1)=1, a(n) = ceiling(r(3)*a(n-1)) where r(3) = (1/2)*(3 + sqrt(13)) is the positive root of X^2 = 3*X + 1.

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A096979 Sum of the areas of the first n+1 Pell triangles.

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A193845 Mirror of the triangle A193844.

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A210197 Triangle of coefficients of polynomials u(n,x) jointly generated with A210198; see the Formula section.

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A293004 Expansion of 2*x^2 / (x^3 + x^2 - 3x + 1).

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A077921 Expansion of (1-x)^(-1)/(1+2*x-x^2).

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A210198 Triangle of coefficients of polynomials v(n,x) jointly generated with A210197; see the Formula section.

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A082574 a(1)=1, a(n) = ceiling(r(3)a(n-1)) where r(3) = (1/2)(3 + sqrt(13)) is the positive root of X^2 = 3*X + 1.